 # Find the area of the region enclosed by the inner Priscilla Johnston 2021-12-20 Answered
Find the area of the region enclosed by the inner loop of the curve.
$r=2+4\mathrm{sin}\theta$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it veiga34
Find the interval in which the inner loop exists.
For this we take $r=0$
$⇒2+4\mathrm{sin}\theta =0$
$⇒\mathrm{sin}\theta =-\frac{12}{}$
$⇒\theta =\frac{7\pi }{6},\frac{11\pi }{6}$
So inner loop lies between the lines $\theta =\frac{7\pi }{6}$ and $\theta =\frac{11\pi }{6}$
Since curve is symmetric about the vertical polar axis. So we canf ind the area of the loop by making double the area between $\theta =\frac{7\pi }{6}$ and $\theta =\frac{3\pi }{2}$
Then Area of the loop is $A=2{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{\frac{12}{r}}^{2}d\theta$
Therefore
$A={\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{\left(2+4\mathrm{sin}\theta \right)}^{2}d\theta$
$A=4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}{\left(1+2\mathrm{sin}\theta \right)}^{2}d\theta$
$=4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}\left(1+2\left(1-\mathrm{cos}2\theta \right)+4\mathrm{sin}\theta \right)d\theta$
$=4{\int }_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}\left(3-2\mathrm{cos}2\theta +4\mathrm{sin}\theta \right)d\theta$
$=4\left[3\theta -\mathrm{sin}2\theta +4{\left(-\mathrm{cos}\theta \right]}_{7\frac{\pi }{6}}^{3\frac{\pi }{2}}$
$=4\left[\frac{9\pi }{2}-\frac{7\pi }{2}-\frac{32}{\sqrt{3}}\right]$
$A=4\pi -6\sqrt{3}$
###### Not exactly what you’re looking for? Archie Jones
The inner loop of the curve is bounded by the 2 points where r=0. Plug that in and you get sin? = -1/2, so ? = 210 and 330 degrees, or 7?/6 and 11?/6. So the integral needed is
$\int \left(2+4\mathrm{sin}\left(\theta \right)\right)d\theta$
Apply the Sum Rule:
$=\int 2d\theta +\int 4\mathrm{sin}\left(\theta \right)dth\eta$ BSJ $\int 2d\theta =2\theta$
$\int 4\mathrm{sin}\left(\theta \right)d\theta =-4\mathrm{cos}\left(\theta \right)$
$2\theta -4\mathrm{cos}\left(\theta \right)$
$=2\theta -4\mathrm{cos}\left(\theta \right)+C$
Compute the boundaries: ${\int }_{\frac{7\pi }{6}}^{\frac{11\pi }{6}}\left(2+4\mathrm{sin}\left(\theta \right)\right)d\theta =\frac{11\pi }{3}-2\cdot \sqrt{3}-\left(2\cdot \sqrt{3}+\frac{7\pi }{3}\right)$
$=\frac{11\pi }{3}-2\cdot \sqrt{3}-\left(2\cdot \sqrt{3}+\frac{7\pi }{3}\right)$
Simplify:
$=\frac{4\pi }{3}-4\sqrt{3}$
###### Not exactly what you’re looking for? RizerMix